题解 | #反转链表#

class Solution {
public:
    ListNode* ReverseList(ListNode* pHead) {
        /*ListNode *pre=NULL;
        while(pHead)
        {
            ListNode *p=pHead->next;
            pHead->next=pre;
            pre=pHead;
            pHead=p;
        }
        return pre;
        */
        if(!pHead||!pHead->next)
            return pHead;
        ListNode *next=pHead->next,*p=ReverseList(next);
        pHead->next=NULL;
        next->next=pHead;
        return p;
    }
};

pre为反转后结点的每个next结点